Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iostream>
#include<queue>
#include<vector>
using namespace std;
#include<vector>
#include<queue>
#include<limits>
template<typename T>
vector<T>dijkstra(int s,const vector<vector<pair<int,T> > >&G,T INF=numeric_limits<T>::max())
{
	int n=G.size();
	vector<T>d(n,INF);
	vector<int>parent(n,-1);
	priority_queue<pair<T,int>,vector<pair<T,int> >,greater<pair<T,int> > >P;
	d[s]=0;
	P.push(make_pair(d[s],s));
	while(!P.empty())
	{
		pair<T,int>p=P.top();P.pop();
		int v=p.second;
		if(d[v]<p.first)continue;
		for(const pair<int,T>&e:G[v])
		{
			int u=e.first;
			T cost=d[v]+e.second;
			if(d[u]>cost)
			{
				d[u]=cost;
				parent[u]=v;
				P.push(make_pair(d[u],u));
			}
		}
	}
	return d;
}
int n,m,s,t;
int main()
{
	cin>>n>>m>>s>>t;
	s--,t--;
	vector<vector<pair<int,int> > >G(n);
	for(int i=0;i<m;i++)
	{
		int a,b;cin>>a>>b;
		a--,b--;
		G[a].push_back(make_pair(b,1));
		G[b].push_back(make_pair(a,1));
	}
	vector<int>ds=dijkstra(s,G);
	vector<int>dt=dijkstra(t,G);
	int lim=ds[t];
	vector<int>cnts(lim,0),cntt(lim,0);
	for(int i=0;i<n;i++)
	{
		if(ds[i]<lim)cnts[ds[i]]++;
		if(dt[i]<lim)cntt[dt[i]]++;
	}
	long ans=0;
	for(int i=0;i<lim-1;i++)
	{
		ans+=1LL*cnts[i]*cntt[lim-i-2];
	}
	cout<<ans<<endl;
}